A) \[f(x)\] is not continuous at x = 0
B) \[f(x)\] is differentiable at x = 0
C) \[f(x)\] is continuous but not differentiable at x = 0
D) None of the above
Correct Answer: C
Solution :
| [c] \[\because f(x)=\left| x \right|+{{x}^{2}}\] |
| \[\Rightarrow f(x)=\left\{ \begin{matrix} {{x}^{2}}+x, & x\ge 0 \\ {{x}^{2}}-x, & x<0 \\ \end{matrix} \right.\] |
| LHL \[=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)\] |
| \[=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,f(0-h)=\underset{h\to 0}{\mathop{\lim }}\,{{(0-h)}^{2}}-(0-h)\] |
| \[=\underset{h\,\to \,0}{\mathop{\lim }}\,{{h}^{2}}+h=0\] |
| and RHL \[=\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0+h)\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,{{(0+h)}^{2}}+(0+h)\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,{{h}^{2}}+h=0\] |
| \[\Rightarrow LHL=RHL=f(0)\] |
| \[\Rightarrow f(x)\] is continuous at \[x=0\] |
| Now, \[LHD=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{-h}\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{2}}+h}{-h}=-\underset{h\to 0}{\mathop{\lim }}\,h+1=-1\] |
| and, \[RHD=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{2}}+h}{h}=\underset{h\to 0}{\mathop{\lim }}\,h+1=1\] |
| Thus, \[LHD\ne RHD\] |
| \[\Rightarrow f(x)\] is not differentiable at \[x=0\] |
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